Difference between revisions of "Mock up midterm exam with solution" - Rhea

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Sample Midterm Examination 2

ECE 438

Fall 2016

Instructor: Prof. Mimi Boutin

(15 pts) 1. List at least three properties of an LTI system.

Solution:

(2)$ y[n] = x[n] * h[n] $

(3)$ {\mathcal y}(\omega) $=$ {\mathcal X}(\omega) $$ {\mathcal H}(\omega) $

(4)$ Y(z) = X(z)H(z) $

(30 pts) 2. For each ROAC, determine which of these system properties apply. (Just list the letters of the properties that apply.) Below we describe the ROAC of the transfer function of an LTI system.

a) the system is causal;

b) the system is BIBO stable;

c) the system has a well defined and finite frequency response function;

d) the system is an FIR filter;

e) The system is an IIR filter;

f) the unit impulse response of the system is right-sided;

g) the unit impulse response of the system is left-sided;

1.1 ROAC= all finite complex numbers, but not infinity.

1.2 ROAC= all complex numbers, including infinity.

1.3 ROAC= all complex numbers z with |z|>0.5, including infinity.

1.4 ROAC= all finite complex numbers z with |z|>3, but not infinity.

1.5 ROAC= all complex numbers z with |z|<0.5.

1.6 ROAC= all complex numbers z with 2<|z|<3.

Solution: 1.1 ROAC= all finite complex numbers, but not infinity. (???????)

1.2 ROAC= all complex numbers, including infinity.

1.3 ROAC= all complex numbers z with |z|>0.5, including infinity.

1.4 ROAC= all finite complex numbers z with |z|>3, but not infinity.

1.5 ROAC= all complex numbers z with |z|<0.5.

1.6 ROAC= all complex numbers z with 2<|z|<3.

(20 pts) 3. Compute the z-transform of the signal

                  $ x[n]= 6^{n}u[n] + 8^{n}u[-n+1]  \  $

Solution:

$ X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n} = \sum_{n=-\infty}^{\infty} (6^{n}u[n]+ 8^{n}u[-n+1]) z^{-n} = \sum_{n=-\infty}^{\infty} 6^{n}u[n] z^{-n} + \sum_{m=-\infty}^{\infty} 8^{m}u[-m+1] z^{-m} $

$ X(z) = \sum_{n=-\infty}^{\infty} (\frac{6}{z})^n u[n] + \sum_{m=-\infty}^{\infty} (\frac{8}{z})^{m}u[-m+1] $

Let k = -m+1, then

$ X(z) = \sum_{n=-\infty}^{\infty} (\frac{6}{z})^n u[n] + \sum_{k=-\infty}^{\infty} (\frac{z}{8})^{k-1}u[k] = \sum_{n=0}^{\infty} (\frac{6}{z})^n + \frac{8}{z} \sum_{k=0}^{\infty} (\frac{z}{8})^{k} $

$ X(z) = \frac{1}{1-\frac{6}{z}} + \frac{8}{z} \frac{1}{1-\frac{z}{8}} = \frac{z}{z-6} + \frac{8}{z} \frac{8}{8-z} , if \quad 6 < |z| < 8 $

$ X(z) = \left\{ \begin{array}{l l} \frac{z}{z-6} + \frac{8}{z} \frac{8}{8-z} &, if \quad 6 < |z| < 8 \\ \text{diverges} &, \quad \text{otherwise} \end{array} \right. $
(20 pts) 4. Compute the inverse z-transform of

                  $ X(z)=\frac{1}{(1+ z)(6-z)}, \text{ ROC }  |z|>6 $

Solution:

$ X(z)=\frac{1}{(1+ z)(6-z)} =\frac{1}{4} (\frac{1}{1+z} + \frac{1}{6-z}) =\frac{1}{4} (\frac{1}{z} \frac{1}{1+\frac{1}{z}} - \frac{1}{z} \frac{1}{1-\frac{6}{z}}) =\frac{1}{4z} (\frac{1}{1-(-\frac{1}{z})} - \frac{1}{1-\frac{6}{z}}) $

So for $ |z| > 6 $, we have

$ X(z)=\frac{1}{4z} [\sum_{k=0}^{\infty} (-\frac{1}{z})^k - \sum_{k=0}^{\infty} (\frac{6}{z})^k] $

$ X(z)=\frac{1}{4} \sum_{k=-\infty}^{\infty} (-1)^k z^{-k-1} u[k] - \frac{1}{4} \sum_{k=-\infty}^{\infty} 6^k z^{-k-1} u[k] $

Let n = k+1, then

$ X(z)=\frac{1}{4} \sum_{k=-\infty}^{\infty} (-1)^{n-1} u[n-1] z^{-n} + \frac{1}{4} \sum_{k=-\infty}^{\infty} 6^{n-1} u[n-1] z^{-n} $

$ x[n]=[-\frac{1}{4} (-1)^n - \frac{1}{12} 6^n] u[n-1] $